An object is thrown upward and reaches a height of 45 meters. It had an initial velocity of 30 m/s. What was the acceleration acting on the object?Given: 𝑠=45 ms=45m, 𝑢=30 m/su=30m/sRequired: 𝑎a (acceleration)
Question
An object is thrown upward and reaches a height of 45 meters. It had an initial velocity of 30 m/s. What was the acceleration acting on the object?Given: 𝑠=45 ms=45m, 𝑢=30 m/su=30m/sRequired: 𝑎a (acceleration)
Solution
To find the acceleration, we can use the second equation of motion which is:
v² = u² + 2as
Where: v = final velocity u = initial velocity a = acceleration s = distance
When the object reaches its maximum height, its final velocity (v) becomes 0. So, the equation becomes:
0 = (30 m/s)² + 2a*45 m
Solving for a, we get:
- (30 m/s)² = 2a*45 m
- 900 m²/s² = 90a m
- a = -900 m²/s² / 90 m
- a = -10 m/s²
So, the acceleration acting on the object is -10 m/s². The negative sign indicates that the acceleration is acting downwards, which is the direction of gravity.
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